Finding arc length of a parametric curve

Formula for arc length of a parametric curve

The arc length of a parametric curve over the interval $\alpha\le{t}\le\beta$ is given by

$L=\int^\beta_\alpha\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\ dt$

where $\alpha$ and $\beta$ are the limits of the interval

where $dx/dt$ is the derivative of $x(t)$

where $dy/dt$ is the derivative of $y(t)$

Example

Find the length of the parametric curve over $0\le{t}\le2\pi$ .

$x=5\sin{t}$

$y=5\cos{t}$

We need to find the derivatives of the parametric equations.

$x=5\sin{t}$

$\frac{dx}{dt}=5\cos{t}$

and

$y=5\cos{t}$

$\frac{dy}{dt}=-5\sin{t}$

Since we were given the limits of integration in the problem, we’re ready to plug everything into the arc length formula.

$L=\int^{2\pi}_0\sqrt{(5\cos{t})^2+(-5\sin{t})^2}\ dt$

$L=\int^{2\pi}_0\sqrt{25\cos^2{t}+25\sin^2{t}}\ dt$

$L=\int^{2\pi}_0\sqrt{25(\cos^2{t}+\sin^2{t})}\ dt$

Since $\cos^2{t}+\sin^2{t}=1$ , we get

$L=\int^{2\pi}_0\sqrt{25(1)}\ dt$

$L=\int^{2\pi}_05\ dt$

$L=5t\big|^{2\pi}_0$

$L=5(2\pi)-5(0)$

$L=10\pi$